Let A(,-2), B(,6) and C(4,-2) be vertices of a ABC.

Question

Let $A(\alpha,-2),\ B(\alpha,6)$ and $C\left(\dfrac{\alpha}4,-2ight)$ be vertices of a $	riangle ABC$. If $\left(5,\dfrac{\alpha}4ight)$ is the circumcentre of $	riangle ABC$, then which of the following is NOT correct about $	riangle ABC$?

Accepted Answer

Circumcentre of $\triangle A B C$

$$
\begin{aligned}
&=\left(\frac{\alpha+\frac{\alpha}{4}}{2}, \frac{6-2}{2}\right) \\
&=\left(\frac{5 \alpha}{8}, 2\right) \\
&=\left(5, \frac{\alpha}{4}\right) \\
&\Rightarrow \alpha=8
\end{aligned}
$$

$\operatorname{area}(\triangle A B C)=\frac{1}{2} \cdot \frac{3 \alpha}{4} \times 8=24$ sq. units

$$
\begin{aligned}
\text { Perimeter } &=8+\frac{3 \alpha}{4}+\sqrt{8^{2}+\left(\frac{3 \alpha}{4}\right)^{2}} \\
&=8+6+10=24
\end{aligned}
$$

Circumradius $=\frac{10}{2}=5$

inradius $(r)=\frac{\Delta}{s}=\frac{24}{12}=2$

Solutions · 1